Round Robin Algorithm Using Circular Linked List - linked-list

Question

Use a circular singly linked list to implement Round Robin process scheduling algorithm in which
each process is provided a fixed time (quantum) to execute and is pre-empted after that time period
to allow the other process to execute. Assume a set of ‘n’ processes are ready for execution.
Read the time quantum and for each of the processes, read the total execution time.
Name the processes as ‘A’, ‘B’ and so on in sequence. Each node should contain the name
of the process, its total execution time and the remaining execution time. If a process
completes its execution, remove it from the list after displaying its name and the
completion time.
Input format:
First line contains the value of ‘n’, the number of processes
Second line contains the time quantum
The remaining lines contain the total execution time of the processes in order.
5
2
6
3
7
5
1
Output:
E 9
B 12
A 18
D 21
C 22

Answer 1

#include <iostream>
using namespace std;
class node
{
public:
char name;
int tm;
int rt;
node *next;
};
class rr
{
public:
node * Head = NULL;
int j = 65;
void insert (int n)
{
node *nn = new node;
nn->name = j++;
nn->tm = n;
nn->rt = nn->tm;
if (Head == NULL)
{
Head = nn;
Head->next = Head;
}
else
{
node *temp = Head;
while (temp->next != Head)
temp = temp->next;
nn->next = temp->next;
temp->next = nn;
}
}
void quantum (int t)
{
node *temp = Head;
int c = 0, i = 0;
while (Head != NULL)
{
{
temp->rt = temp->rt - t;
c = c + t;
if (temp->rt <= 0)
{
c = c + temp->rt;
cout << temp->name;
cout << c << endl;
del (temp->name);
if (temp->next == temp)
{
break;
}
}
temp = temp->next;
}
}
}
void del (char x)
{
node *p = NULL;
node *temp = Head;
if (Head->name == x)
{
while (temp->next != Head)
temp = temp->next;
p = Head;
temp->next = Head->next;
Head = Head->next;
delete p;
}
else
{
while (temp->name != x)
{
p = temp;
temp = temp->next;
}
p->next = temp->next;
delete temp;
}
}
};
int
main ()
{
rr robin;
int i, n, x, y, t;
cin >> y;
cin >> t;
for (i = 0; i < y; i++)
{
cin >> n;
robin.insert (n);
}
robin.quantum (t);
return 0;
}